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Prove $\vec{\nabla} \cdot\left ( \vec{A}\times\vec{B} \right )=\vec{B}\cdot\left ( \nabla \times\vec{A} \right )-\vec{A}\cdot\left ( \nabla \times\vec{B} \right )$

I have expanded the LHS for this and obtain a horrible expression that would be even tedious to put them into latex form without a headache.

I attempted to rearrange the LHS so as to obtain the RHS but to no avail.

any help is appreciated.

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    $\begingroup$ My comment is at a slightly more advanced level but I will write it anyway. Think of $X=\mathbb{R}^3$ with its flat Euclidean metric. I will use the d notation. If you have a function $f$ on $X$, $df$ is essentially the gradient of $f$. If you have a $1$-form $\alpha$, $d\alpha$ is essentially the curl of the vector field obtained by $\alpha$ by raising its index. If $\beta$ is a $2$-form, then $d\beta$ is the divergence of the corresponding vector field. I am implicitly using the Hodge $*$ duality between $1$-forms and $2$-forms, as well as between functions and $3$-forms, on $\mathbb{R}^3$. $\endgroup$
    – Malkoun
    Mar 19, 2017 at 20:20
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    $\begingroup$ Continuing my previous comment, the identity you asked about corresponds simply to $d(\alpha \wedge \beta) = d\alpha \wedge \beta - \alpha \wedge d\beta$, where $\alpha$ and $\beta$ are $1$-forms on $\mathbb{R}^3$. $\endgroup$
    – Malkoun
    Mar 19, 2017 at 20:22

3 Answers 3

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In index notation

\begin{eqnarray*} (A \times B)_{i} = \epsilon_{ijk} A_j B_k \end{eqnarray*} (Einstein's convention of sum over repeated indices). Then if $A_{j_{,i}}=\partial A_j/\partial x_i$, and from $\nabla \times A=\epsilon_{ijk} A_{k_{,j}}$ (and so for the other symbols)

\begin{eqnarray*} \nabla \cdot (A \times B) &=& [\epsilon_{ijk} A_j B_k],_{i} \\ &=& \epsilon_{ijk} A_{j_{,i}} B_k + \epsilon_{ijk} A_j B_{k_{,i}} \\ &=& B_k ( \epsilon_{kij} A_{j_{,i}}) - A_j ( \epsilon_{jik} B_{k_{,i}}) \\ &=& B \cdot (\nabla \times A) - A \cdot ( \nabla \times B ) \end{eqnarray*} where the minus ``-'' sign appears since $\epsilon_{ijk}=-\epsilon_{jik}$.

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  • $\begingroup$ This is more convenient. $\endgroup$ Dec 16, 2020 at 5:41
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You just need to write everything out neatly, so you can see the equating terms.

Let's deal with the left hand side first. $$\vec{A}\times\vec{B}=(A_2B_3-A_3B_2)\vec{i}+(A_3B_1-A_1B_3)\vec{j}+(A_1B_2-A_2B_1)\vec{k}$$ I wrote the component of $A$ always in front of the component of $B$ in order to see easily. Now the whole left hand side is the divergence of the above expression, and therefore equal to: $$\frac{\partial(A_2B_3-A_3B_2)}{\partial x}+\frac{\partial(A_3B_1-A_1B_3)}{\partial y}+\frac{\partial(A_1B_2-A_2B_1)}{\partial z}$$

Let's wait for a while to do the product rule, and instead, look at the right hand side. $$\nabla \times \vec{A}=(\frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z})\vec{i}+(\frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial x})\vec{j}+(\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y})\vec{k}$$ The first term in right hand side is the dot product of $B$ with it, so just replace the $i,j,k$ by $B_1,B_2,B_3$: $$\vec{B}\cdot \nabla \times \vec{A}=B_1(\frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z})+B_2(\frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial x})+B_3(\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y}) $$

Now the second term in the right hand side is similar. We just need to switch $A$ and $B$, and remember it is negative: $$\vec{A}\cdot \nabla \times \vec{B}=A_1(\frac{\partial B_3}{\partial y}-\frac{\partial B_2}{\partial z})+A_2(\frac{\partial B_1}{\partial z}-\frac{\partial B_3}{\partial x})+A_3(\frac{\partial B_2}{\partial x}-\frac{\partial B_1}{\partial y}) $$

Now you can compare each product in the left hand side with the corresponding terms in the right hand side. For example, the first product rule in the left hand side is $$\frac{\partial(A_2B_3)}{\partial x}=A_2\frac{\partial(B_3)}{\partial x}+\frac{\partial(A_2)}{\partial x}B_3$$

It is easy to see which terms in the right hand side equal to these two terms. I'll leave it to you to proceed.

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  • $\begingroup$ You deserve a medal $\endgroup$ Jan 11, 2016 at 13:15
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In the language of differential forms:

Firstly understand the wedge product discussed in here, then notice the following correspondance:

$$d(\alpha \wedge \beta) <-> \nabla \cdot( a \times b)$$

Where $\alpha$ and $\beta$ are both one forms, now by the product rule for forms:

$$d(\alpha \wedge \beta)= d \alpha \wedge \beta+(-1)^p \alpha \wedge d \beta$$

Now, note that following points:

  1. There exists another correspondence $d \alpha \to \nabla \times \alpha$

  2. Note that $ \alpha \wedge \beta$ corresponds to Euclidean dot product when $\alpha$ is a one form and $\beta$ is a 2 form

  3. $p=1$ since $\alpha$ is a one form

We get the following identity when translating the product rule of forms into that of vector caclulus:

$$\nabla \cdot(a \times b ) = (\nabla \times a) \cdot b - a\cdot (\nabla \times b) $$

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  • $\begingroup$ Wonder why this got downvoted $\endgroup$ Feb 11, 2022 at 20:45

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